B

Step 1: Understand the configuration of the battery in both cases. In the first configuration, we have
- m rows in parallel, each containing n cells in series.
Step 2: The total voltage of the first configuration can be expressed as
$$ V_1 = nV $$
where V is the voltage of each cell. The equivalent resistance for one row of n cells in series is given by
$$ R_s = nR $$
where R is the internal resistance of each cell. The total resistance when m rows are in parallel is:
$$ R_p = \frac{R_s}{m} = \frac{nR}{m} $$
Step 3: The maximum current I through the external resistor R_{ext} can then be calculated using Ohm's law
$$ I = \frac{V_1}{R_{ext} + R_p} = \frac{nV}{R_{ext} + \frac{nR}{m}} $$
Step 4: In the second configuration, we switch the arrangements to
- n rows in parallel, each containing m cells in series.
The total voltage in this case is
$$ V_2 = mV $$
The equivalent resistance of one row of m cells in series is
$$ R_s' = mR $$
Hence the total resistance is:
$$ R_p' = \frac{R_s'}{n} = \frac{mR}{n} $$
Step 5: The currents in the external resistor for the new configuration is:
$$ I' = \frac{V_2}{R_{ext} + R_p'} = \frac{mV}{R_{ext} + \frac{mR}{n}} $$
Step 6: To find the relation between I and I', we can substitute the expressions we derived. It can be shown that
$$ I' = \frac{mV}{R_{ext} + \frac{mR}{n}} = \frac{mnR + mRV}{R_{ext}(n+mR)} I $$
Therefore, the current through the same external resistor with the new configuration is:
$$ I' = \frac{nm}{m+n} I $$
The options might be like ratio between the currents.
Hence, the answer corresponds to option B: